9.5. Alternating series. Absolute convergence and conditional convergence

Transcription

1 Chpter 9: Ifiite Series I this Chpter we will be studyig ifiite series, which is just other me for ifiite sums. You hve studied some of these i the pst whe you looked t ifiite geometric sums of the form: 3. Does it rig bell? I this Chpter, we cosider more geerl poit of view, where the mi questio will be to determie which ifiite sums dd up to somethig fiite d which do ot (which series coverge d which ot, similr with the defiitios for improper itegrls). To swer to this questio, we will develop series of tests, the collectio of which should cover my types of series. I this sese, studyig series is like studyig itegrls, we use certi types of tests for certi series, d importt prt of solvig problem is to determie which test to use for which series. Sice y series is sequece, we will begi the chpter with reviewig sequeces. TABLE OF CONTENTS: 9.. Ifiite sequeces 9.. Itroductio to ifiite series 9.3. Positive series: The itegrl Test 9.4. Positive series: Other Tests 9.5. Altertig series. Absolute covergece d coditiol covergece 9.6. Power Series 9.7. Opertios o Power Series 9.8. Tylor d McLuri Series 9.9. The Tylor Approimtio Theorem

2 9.. Ifiite sequeces: { } Defiitio: A sequece, is ordered (ideed) list of umbers. Wys to defie sequece: list the terms of sequece. Emple: (),4,9,4, ; give formul (defiitio) for the geerl term of the sequece: () 5( ) 5 6, ; This is the preferred method sice it is so esy to use (plot, clculte limit, etc.) d we ofte eed to obti formul like () from the other forms (() or (3)); give recurrece reltio d the iitil term: (3) ; 5, ; or the Fibocci f ; f ; f f f, 3 sequece: Emples of sequeces:,, si( ), Note tht does t eed to strt t i the defiitio of sequece. NOTE: Note tht y sequece c be viewed s prticulr fuctio f : N R, f ( ). Therefore, whe plottig sequece, we plot versus, s plottig fuctio. The most importt questio for sequece is its covergece. Defiitio: A sequece coverges to rel umber L, d we deote this by lim Lif 0, such tht, N N L N. (This is the defiitio for the covergece of fuctio t, tht is the terms of the sequece get s close s we wt to L if is big eough). A sequece which is ot coverget is diverget. Emples:, ;, ; si( ), ;, ;, ;, -- Plot terms to guess L d prove L by defiitio.

3 More efficiet methods to clculte limits re discussed below. NOTE: If lim, we sy tht diverges to. (Keep this is mid). NOTE: A sequece is coverget if it hs fiite limit t ifiity. NOTE: Sice y sequece is fuctio, whe studyig the limit of sequece we relly study the limit of fuctio t. Therefore, ll previous tools which we hve for clcultig the limit of fuctio (limits of elemetry fuctios, opertios with limits, fudmetl limits, d specil form of l Hospitl s Rule (which will be preseted soo) c be used). To clculte the limit of sequece we ll use: --- limits of elemetry fuctios (polyomils, rtiols, epoetil, etc) / epressios, opertios with limits; --- fudmetl limits, i.e. 0, 0 ; --- dpted form of l Hospitl rule see below; --- squeeze theorem; --- mootoic sequece theorem. Let us develop ech of these. The followig rules re true: Theorem : If d b re coverget sequeces d k is (rel) costt, the: lim k k lim k k lim lim b lim lim b lim b lim lim b lim lim, if lim b 0 b lim b () Emples: Clculte: ) lim 3 5 b) lim c) lim l( ) 3

4 I the bove emple c), we cot pply l Hospitl s Rule directly (cot tke the derivtive of sequece discrete set of poits --). Isted, we use the followig: Theorem : If lim f ( ) L the lim f ( ) lim L. Proof: The result is obvious (if cotiuous fuctio coverges, the the sequece which represets the fuctio discretely coverges to the sme limit). It c be proved by defiitio. Therefore, whe we eed to use l Hospitl s Rule i sequece, ALWAYS trsform it first to fuctio: lh ' l( ) Th. l( ) lim lim lim lim. Never tke the derivtive of sequece, but pply the method bove (otherwise poits will be deducted ). Aother useful result to clculte limits of sequeces: Theorem 3 (squeeze theorem) : Let lim L d lim c L d let d c be coverget sequeces to the sme limit L such tht b, c K, where K is rbitrry turl umber. The lim b () L. The theorem bove is represeted grphiclly : The squeeze theorem is used ofte with trigoometric fuctios si( ) d cos( ) which c be bouded, but cot be evluted directly t. Emple: si( ) lim Plot, b d c o your clcultor i this cse. 4

5 Theorem 4 (mootoic sequece theorem): If is icresig sequece ( ( such tht, K ) d bouded bove, U U K ), the Similrly, If is decresig sequece ( ( such tht, B B K ), the coverges to L U. K ) d bouded below, coverges to L B. Ituitively this theorem should mke sese. I the first cse, sice the sequece is icresig d ot goig to, it should pproch fiite limit. The proof is bsed o showig tht coverges to the lest upper boud of S Emples:, usig the properties tht L is upper boud, but L is ot. Determie if the followig sequeces re coverget, d if they re, clculte lim : ) b) Hit: Show tht! 3!! is icresig d tht 3 ( ) (Problem 34) b Show tht b b by formig perfect squre d b 0. ;, bouded bove (which we lso show, tht is ). c). List few terms to guess mootoicity, show tht bouded bove (which we lso show, tht is 6). d) ; 6,. List few terms to guess mootoicity, show tht is icresig if is icresig if These methods (mootoic sequece theorem d squeeze theorem) re the oly fesible methods to use for these types of problems (recurret sequeces d trigoometric sequeces) so ler them well. Homework: Wht left from bove plus problems,,4,6,9,7,3,5,3,33,36,37,45. E.C. 50, 53 E.C.: Fid the limit of:,,, 5

6 9.. Itroductio to ifiite series: Defiitio: A ifiite series (sigulr sme s plurl) is ifiite sum. Give sequece, the ifiite series is deoted by. We hve studied some fiite sums before: ( ), 4, et. l, but ow we wt to look t ifiite sums. For emple, wht would be? Wht bout 4 Ifiite series hve my res of pplictio: Deciml represettio of umbers; Emple: does it work out? Do the sme for 9, 3. The defiitio of the defiite itegrl: f ( ) d lim f b k k k k Represettio of lyticl fuctios s power series : i your clcultor most fuctios which re ot polyomil (e.g., e, l( ) ) re represeted first s ifiite polyomils (power series) before they re evluted. We will see how this hppes i the fil sectios of this chpter. Solutio of some differetil equtios: lyticl series method for lier or olier differetil equtios (Cuchy) or geerl theory for o-lier differetil equtios roud criticl poits. Like we sw bove, some series diverge (to ), while others my coverge to fiite vlue. This issue of covergece for series is the most importt i the study of series d cetrl to this chpter. 6

7 I order to study the covergece of ifiite series, it is most coveiet to cosider the prtil sums. Note tht the prtil sums form sequece i themselvess S k k sequece coverge, ll we hve to do is to clculte lim s.. To see if Defiitio: The ifiite series is clled coverget if the sequece of prtil sums S coverges (to rel vlue), tht is if lim S S. Otherwise (if lim S or if lim S does ot eist), the series is diverget. Notice tht = lim S lim k. For coverget series, S lim S is clled the sum of the series. k Emples: Is coverget? Wht bout? I geerl, ifiite geometric series is series of the form r r r r We kow tht r k S r r r r. r k Therefore,, if r lim S r or D.N.E. if r. This is geerl result tht we should memorize: A ifiite geometric series of the form r r r is coverget to whe the bsolute rtio r, d is diverget r otherwise. Emples: coclude directly bout , , 7

8 , 3 Telescopic series: They re series (usully formed from frctios) i which the terms ccel cosecutively if writte ppropritely. Emple:, ( ) S A B k k( k ) k k k, we fid tht A d B such tht S, d k k k therefore is coverget to. For simple frctios like bove, it is worthwhile to see if it is ( ) telescopig series. Theorem (The Divergece Test --- th Term Divergece Test): If geerl series is coverget, the lim 0. Coversely, if lim 0, the is diverget. NOTE: This test (the first geerl test tht we hve) is very importt d it will help us idetify my diverget series. However, keep i mid tht this test is useful to recogize diverget series, d ot coverget series (it is coclusive oly whe lim 0 ( D), but icoclusive whe lim 0 ), i.e. the coditio lim 0 is oly NECESSARY coditio for covergece, d NOT A SUFFICIENT coditio. Emples: Try to pply the divergece test (or other tools) to study the covergece of the followig series:,,, l( ), l( ), For the lst two series, the divergece test is icoclusive. Actully, we c show tht (D). ( C) d tht For, show tht S by cosiderig S, S4, S8, etc.. will be studied bit lter. 8

9 Theorem (opertios with series): If d b re coverget series, the so re the series c, b d b d c c, b b, b b. NOTE: The bove theorem does t sy ythig bout the cse whe oe of the series (or both) re diverget. The cse d b ( C) d (D) is hrder. b (D) usully implies tht b (D). Why? However, the cse (D) Emples: 3 ( ), 3 ( ),. Homework: Wht left from bove, plus,,5,6,7,9,,8,,3,6. E.C.: 9,34,50 9

10 9.3. Positive series: The itegrl Test: The focus of the et sectios is to determie the covergece of series, lthough occsiolly we will be ble to determie estimtes for the sum of some series too. We will derive series of tests (of the type of the divergece test preseted i 9.) which we ll help us determie immeditely if series is coverget or diverget. For the et two sectios, we restrict to the study of positive series oly (series for which most terms re positive i.e. with 0, for K ). This is becuse it is much esier to devise covergece tests for positive series. However, do ot forget this restrictio, d lwys check tht the series is positive series before pplyig y of these tests! Emple: Which of the followig re positive series?,, si( ),,, cos( ) The oly dger of divergece for positive series is tht they c dd up to, sice the prtil sums re lwys icresig. You c red the Bouded Sum Test which discusses tht. Optiol: Lemm (Bouded Sum Test): A positive series bove. is coverget if d oly the sequece of its prtil sums S is bouded from k k Proof: : Note tht S is icresig, sice S S 0 (positive series). If S the due to the mootoic sequece theorem, it is coverget. : If this implies tht S is bouded. is upper bouded, is coverget, the S is coverget to fiite vlue S. Usig the defiitio of covergece, NOTE: The result bove is Lemm, d we will use it idepedetly oly rrely. However, it is prmout for positive series d is used i the followig theorem. Emple:! Ed Optiol 0

11 Theorem (The Itegrl Test for Positive Series): Cosider the positive series d the correspodig fuctio :[, ) [0, ) ssumed to be cotiuous, positive d decresig fuctio ( f, f ( ), which is f '( ) 0, [, )). The the ifiite series coverges if d oly if the improper itegrl f ( ) d coverges. NOTE: This is very importt result, which reduces the study of the covergece of series to the covergece of improper itegrl (much esier to evlute). However, before usig it, you SHOULD ALWAYS CHECK THAT the fuctio f() is cotiuous, positive d especilly DECREASING over its domi. NOTE: The itegrl test is used whe ll coditios bove re met (d reltively esy to check), d whe we re ble to clculte reltively esily f ( ) d. NOTE: The itegrl test is ecessry d sufficiet coditio (it goes both wys): therefore, it c be used for both coverget d diverget series. Emples: Use the itegrl test bove (or other test) to determie the covergece of the followig series:, rct( ),, 0, si( ),,,,! si Let us look i geerl t the p-series: p, 0 p for ow. f( ) p p p f '( ) p p 0 over the fuctio s domi, so ( ) The fuctio f :[, ) [0, ), itegrl test re met. is positive d cotiuous. Also, f is decresig. All coditios of the

12 , whe p 0 p p p p d, whe p p. Therefore,, whe 0< p ( C) for p p is ( D) for p. (you could lso see directly tht p (D) for 0 p -- why?). This is importt result for series which we should memorize. Combied with some other tests/et sectio, we should be ble to determie the covergece of my ew series, d especilly of lmost y series produced by rtiol d irrtiol fuctios! Emples: Determie the covergece of the followig series:,, 00 4,,, e,, 3 5,, 4 3, l( ) e,. The Riem zet-fuctio is defied by ( ) d is used i umber theory to study the distributio of prime umbers. Wht is the domi of? Proof (of the itegrl test): Figure 9.. The fuctio f ( ) d the sequece

13 Cosider decresig positive cotiuous fuctio f :[, ) [0, ), give by f ( ). Lookig t picture bove, it is esy to see tht () 3 f ( ) d 3 Therefore, () f ( ) d k k k k. If f ( ) d is coverget, the f ( ) d U (this sequece is coverget, d therefore it is upper bouded, sice icresig), so from the iequlity () we hve tht k (upper bouded). Sice S S 0 k S U sequece theorem, it follows tht S ( is coverget. Q.E.D. Coversely, if bouded (sice icresig), Therefore, from () it follows tht the sequece icresig), so by the mootoic sequece theorem, A estimte for the sum of series: f ( ) d (d therefore S is icresig), from the mootoic ( C) S S is upper f ( ) d is upper bouded (d f ( ) d ) is coverget. Q.E.D. The tests before oly determie if series is coverget or diverget, but how c we determie (or t lest estimte) the sum of series. For positive series, d whe the coditios of the itegrl test re met, we c fid estimte for S with its prtil sum S : S, which represets the error tht we mke by pproimtig the sum of the series 3

14 Figure 9.. Estimte of the error S S usig the itegrl test From Figure 9..., we see tht: S S S S f ( ) d, which gives us very useful estimte for the error S S. Emple: Give estimte of the error mde by pproimtig the sum of the series with tkig prtil sum with 0 terms. (Compre with the ect vlue of the sum: 6 ). How my terms should we tke i the prtil sum to hve error less th ? Homework: Wht left from bove, plus,3,4,8,,3,5,8,9,4,5,6,9,30,33,34. E.C.: 35,36,37 4

15 9.4. Positive series: Other Tests: The itegrl test is very useful test, but it requires the verifictio of few coditios, d sometimes it cot be pplied. Let us look t the series: effort.. This series c be lyzed with the itegrl test, but tht will require some However, this series looks similr with, with which we d like to compre it. Theorem (Ordiry Compriso Test): Cosider 0 b, for K. The: ) If b) If b ( C) the ( D) the ( C). b ( D) I words, positive series smller th coverget series is coverget, greter th diverget series is diverget. Relte this to the observtio bout the dger of positive series to diverge to ifiity whe they diverge. The proof for ) is esy usig bouded prtil sums d mootoic sequece theorem. b) follows from ) usig cotrdictio. For the emple bove:, so ( C). Other emples: si ( ),, 5,,,, , (some of these my be hrder th they seem d my be solved much esier with the limit compriso test below). Usig the ordiry compriso test d the p-series, we should be ble to quickly prove the covergece/divergece of ll rtiol type series d of my other series. However, the ordiry compriso test stops short of series whe oly sig chge hppes, i series like. For this, we give d we ll use the stroger test: 5

16 Theorem (Limit Compriso Test): Cosider 0 d b 0 d let () lim b L. If 0 L, the the series d b hve the sme behvior! (very strog); If L=0 (thik of this like b) the if b ( C) the ( C) ; If L (thik of this like b) the if b ( D) the ( D). The proof for ) uses the defiitio of the limit with some ϵ d Theorem fter. Similrly for b) d c). Remrk: This is esier to use test (be creful with the coditios) d it c be pplied for lmost ll emples d type of problems which require compriso test. Eceptio for the give problems is si ( ). Other emples: emple). l( ) l( ),,, 9 (this lst c be compred with for! (keep i mid tht l() is smller (slower) th y power > 0 of ). A eve esier to use test for positive series is: Theorem 3 (Rtio Test): Let 0 d ssume tht () lim. The: If, the If, the ( C). ( D). If, the the test is icoclusive. Ituitively, this theorem mkes sese whe we thik of () to sy tht is geometric series with rtio ρ, for which the bove coclusios re true. The proof follows these lies. The rtio test is pplied whe there re my simplifictios i d you epect lim. 6

17 Emples:,, ( ),!,!!, 3 4 cos( ), -- use ordiry compriso test 3 ( )! Homework: Wht left from bove, plus,,6,7,0,,3,6,9,,7,3. E.C.: 43,c, 44, Altertig series. Absolute covergece d coditiol covergece: Most of the previous tests pply to positive series oly. I order to eted our study to series with rbitrry terms (positive d egtive), we begi by studyig i this sectio specil type of o-positive series, mely ltertig series. Altertig series re series whose terms lterte sigs. Emples: 3 4 ; A geerl ltertig series is series of the form : () b, where 0, b K. I () we usully ssume tht the first term is positive, if ot, we c fctor sig from the etire series. Theorem (Altertig Series Test): Cosider the ltertig series b, with 0, ) is decresig, tht is, b K. If : b b b K d b) lim b 0 The b is coverget. I this cse, estimte for the error S S is S S b. NOTE: The most importt (d lmost oly coditio) to check for ltertig series is tht b decresig. Coditio b) is merely ecessry coditio (remember divergece test, which hs to be verified lso. We symbolize i short form the two coditios s 0 b. Note tht the Altertig series test is very esy test to do, but it is sufficiet test oly, d ot ecessry (it is icoclusive whe the coditios bove re NOT MET) to be 7

18 Note the specil form of ltertig series, which should be verified before you use the test. Proof: Figure Terms of S for ltertig series with b 0 The terms of the prtil sum S for ltertig series uder the coditios of the theorem re sketched I Figure Note tht S3 b ( b b3 ) S b sice b b3 0 I geerl, it is esily show tht S S (S S b ( b b ) ( b b ) ( b b ) b (sice b is decresig). Similrly, S4 S is icresig) d. Therefore S S S b S, sice b 0. is coverget. From the picture bove, it is cler tht S S S (or the other wy roud), so S S S S S S b. Q.E.D. Emples: Study the covergece of. How my terms do we eed to tke i S such tht SS 0.00?. Study the covergece of. How close is S 5 to S?! The test bove is esy to use, but quite limited. It does ot sy ythig bout geerl o-positive series (ot ecessrily ltertig), or bout the cse whe the coditios re ot met. si( ) Emple: How do we study the series? I order to del with geerl series like the series bove, we itroduce few otios. 8

19 Defiitio: A geerl (possibly o-positive) series coverges. is sid to coverge bsolutely (AC), if (remember this s coverget i bsolute vlue). Note tht the secod series is much esier to study, sice it is positive series. Theorem (bsolute covergece): If is bsolutely coverget (AC), the is coverget ( C). Proof: Cosider v v. Sice 0v, d sice Therefore is ( C), the v ( C). ( C) (differece of two coverget series). Theorem is very useful whe it is ot is bsolutely coverget ( si( ) ), but it does ot cover the cses whe ( si( ) ). Of course, there re series which re ot bsolutely coverget, but just coverget. Defiitio: If (CC). is coverget but is diverget, the we sy tht coditiolly coverget Emples: Study if ech of the followig series is (AC) or (CC). Hit: It is usully good to strt with the covergece of (AC), to void uecessry work).,,!,, si(4 ), 4 cos( ), cos 3! The cses i which the series i ot bsolutely coverget ( is diverget), but the series is ot ltertig, re the hrdest cses to study if the series is (CC) or (D). The followig theorem c sometimes come to rescue (it is much stroger th it seems). Ad it will be useful i the et sectios lso. Theorem 3 (Absolute Rtio Test): 9

20 Let be rbitrry series (ot ecessrily positive). Cosider lim. The: If the the series coverges bsolutely (this is just the rtio test for If, diverges (very useful d ot obvious). If the the test is icoclusive. The proof relies o lim 0 lim 0. ) so coverges. We will use Absolute Rtio Test sometimes for covergece, but be wre tht whe studyig the covergece of, you re ot limited t this test! Emples: Clssify ech of the followig series s (AC), (CC) or (D): si( ), si, si (use AST for C),, 4 (use AST for C) l( ),, e ( ) l( ) Homework: From bove plus,,3,6,7,3,4,5,8,3,6,30,3,40 This is good plce for test. 0

21 9.6. Power Series: I the previous sectios of this Chpter, we hve covered the importt theory of series of umbers, tht is series of the form, where is positive or rbitrry o-positive sequece. For the remiig sectios, we cosider more geerl type of series: power series, which re series of the form: () I (), we covee tht, eve for =0. Note tht the power series () reduces to usul series of umbers for ech fied vlue of. Also, ote tht i most cses the power series () is o-positive, sice i geerl c hve egtive vlues (d itself c be o-positive). The first questio tht we pose for power series is: For wht vlues of is the power series () coverget? Whe coverget, the sum of power series of the form () is fuctio f( ), sice for every 0 where the series is coverget, the series coverges to distict vlue f( 0). We write this i the form: () f ( ) I order to fid the vlues of where power series coverges, we lwys use the bsolute rtio test. Emple : For wht vlues of re the followig power series coverget? 0! Clculte b ( )! lim lim lim lim( ). b! Therefore, whe 0,, d! (D) (see the bsolute rtio test). 0 Whe 0 (tht is 0 ), the series is cosidered 0 0!, d the series coverges. 0 Therefore! for 0 0 d diverges otherwise.

22 The series 0 is clled power series cetered t 0. Most geerlly, we cosider power series cetered. 0 t some vlue of the form: (3) Emple : For wht vlues of re the followig power series coverget? 0 3 Applyig the bsolute rtio test, we fid tht 3. Therefore, is ( C) for d (D) for 3 (,) (4, ). I the edpoits = d =4 the rtio test is icoclusive, d we study these cses seprtely, usig the usul theory of series of umbers: 3 =: =4: 3 ( C ) 0 0 (use the ltertig series test) 3 ( D ) 0 0 (the hrmoic series). Therefore, the series 0 coverges for [,4) d diverges otherwise. 3 Theorem : For geerl power series The series The series coverges oly for 0 coverges for ll R 0 there re three distict possibilities: 0 d diverges otherwise (see emple ); ; There is umber R (clled rdius of covergece), such tht the series coverges for R (possibly icludig oe or both edpoits), d diverges for R (see emple 3).

23 Proof: For geerl power series (3), pply the bsolute rtio test to determie the itervl of covergece: b lim lim lim b why below).. Deote lim R / lim lim (see R The coclusio follows from lyzig the three possibilities: lim 0 R (the series coverges everywhere); lim R 0 (the series coverges oly for ); lim v R / v (the series coverges o R, R possibly icludig the edpoits). NOTE: You c use directly the formul bsolutely rtio test every time for clrity. R lim, lthough it is preferbly to pply completely the NOTE: The itervl R, R (possibly icludig oe or both edpoits) is clled the itervl of covergece of the power series. The first prt of studyig power series is to determie its itervl (rdius) of covergece. The fuctio to which power series covergece is studied fter (d we will tckle tht i the followig sectios). Emples: Solve problems 8, 3,5,6, i the tet Homework: From bove, plus problems 4,8,9,,6,,3,5,7,3,34,35. E.C. 36 3

24 9.7. Opertios o Power Series: Sice we kow how to fid the itervl of covergece for geerl power series to swer the other importt questio for power series: 0, ow we wt () Wht is the fuctio f( ) which is the sum of power series over its itervl of covergece? (ote tht we metioed this t the begiig of the chpter whe we sid tht lyticl fuctios i your clcultor re epressed first s power series before beig evluted!). I order to swer this questio, we usully strt with the opposite simpler questio: () Hvig lyticl (tht is with ifiitely my derivtives over its domi) fuctio, c we write this fuctio s power series? Aswerig this questio will help us (hopefully) fid the sum of power series by recogizig it s epsio of certi fuctio (more o this lter). I this d the et sectio we give methods to swer questio (). For my power series it is useful to strt with the geometric series: (3), for (the restrictio comes from the usul study of geometric series such tht 0 lim S lim R). Covice yourself tht this grees with the itervl of covergece of the power series 0 (bsolute rtio test)! Formul (3) c be regrded s the fudmetl power series. I this sectio, we show how to derive my ew power series bsed o the formul (3). New power series c be derived from (3) bsed o trsformtios of this (or combitios of them):. Substitutio: Sice (3) is vlid for ANY R with, the the vrible i this formul c be replced with y ew epressio (or vrible) s eeded. For emple, we derive: for (4) 0 (5) for 0 4

25 (6) b b for b for y, b R d so o. 0. Wht bout fidig power series for f( )? We could use f ( ) d deduce geerl formul, but is there simpler wy? Yes, we c tke the derivtive of the formul (3) NOTE: If ( ) f hs power series epsio 0 R, R (possibly with edpoits), the '( ) o some covergece itervl o f hs the power series epsio R, R (possibly with edpoits). A similr results holds for itegrtio of (3). This geerl result is R, 0 is ot very hrd to prove: just swer the questio: If the rdius of covergece of d d? Similrly for itegrtio). 0 wht is the rdius of covergece of Tkig the derivtive of (3), we obti: (7) d 3 ' 3 for d (the derivtive of the series but with moved over by sice the derivtive of the first term is 0). (It is much clerer to tke the derivtive i the epded form, t lest for the begiig). Note tht we kow hed of time tht the itervl of covergece for the series i (7) is the sme with the series i (3). Why? 3. Itegrtio: C we fid power series epsio for f ( ) l( ). Note tht the domi of f( ) is,. We remrk tht f d d d 3 3 (8) ( ) l for. Mke the substitutio y i (8) to get: 3 0 5

26 (9) y ( ) y ( ) y for y y? l( y) Therefore we hve derived power series for the fuctio l. I similr mer, we c derive power series for other importt fuctio. 4. e : Cosider the power series 3 S ( ) You c esily fid tht the itervl of covergece!! 3! for this power series is. Also, ote tht S '( ) S( ), for ll. Solvig this differetil equtio, with S(0) (remember Sectio 6.5), we obti S( ) e. Therefore, we hve foud tht: (0) 3 S ( ) for ll.!! 3! You should memorize some essetil epsios (for 5. Other opertios: e,, but the method is more importt). We c derive power series by dditio, subtrctio, multiplictio, (log) divisio, of kow power series. Emples: Fid the power series epsio of : ) f( ) b) 4 f ( ) e c) rct( ) f( ) d) e f ( ) 3 e) f ( ) 3 Homework: Wht left from bove, plus,3,4,5,6,8,5,6. E.C.: 33,36-hrder 6

27 9.8. Tylor d McLuri Series: We cosider ow the more geerl problem: Give lyticl (with ifiitely my derivtives over its domi) fuctio, c we fid power series epsio for this fuctio roud poit i its domi. First, let us ssume tht power series eists for give fuctio. Wht is this power series? Theorem (uiqueess theorem): Let f : D R be lyticl fuctio over its domi D d let D rbitrry poit. Assume tht f( ) is epded s power series roud : The : 3 () f ( ) c c c c o D 0 3 () c ( f ) ( ), where! ( f ) ( ) deotes the th derivtive of the fuctio f( ) evluted for. (covetio f (0) ( ) f ( ) ). NOTE: The bove theorem sys tht is fuctio f() c be represeted s power series, this power series is uique (the coefficiets re uiquely determied by f). Proof: Let us prove () (let us fid the coefficiets of the power series), is we ssume (). Evlute () for to get: f ( ) c0. Tke the derivtive of () to get: 3, d evlute for to get f '( ) c, which gives c (3) f '( ) c c 3c (grees with ()). Tke the derivtive of (3) to get: f ''( ) c 3c3 4 3c4 d evlute for to get f ''( ) c c (4) f ''( )! Cotiuig this procedure it is cler tht we obti the formul () for the coefficiets c i the power epsio of f( ) roud (this power series is clled the Tylor epsio or the Tylor series of f( ) roud. A more rigorous proof c use mthemticl iductio to show (). Q.E.D. 7

28 Therefore, the geerl Tylor series of fuctio f( ) roud is: f f f f!! 3!! (3) ( ) '( ) ''( ) ( ) ( ) 3 (5) f ( ) f ( ) Formul (5) is very importt formul which eeds to be memorized (we ll use it ofte). Defiitio: The Tylor series of fuctio f( ) roud 0 is clled the McLuri series (epsio) of f( ). It hs the form: 0 (6) (3) ( ) f '(0) f ''(0) f (0) 3 f (0) f ( ) f (0).!! 3!! 0 Emples: Fid the McLuri series for f ( ) e, f ( ) l( ), f ( ) si( ), f ( ) cos( ). Do these gree with the formuls you hve foud before? There is still questio bout if the Tylor series eists for give fuctio (lthough we kow wht this series is, whe it eists). We hve hited tht ll lyticl fuctios (fuctios which hve ifiitely my derivtives) dmit Tylor series. The importt Tylor s theorem et sttes this: Theorem (Tylor s theorem): Let f : D R be fuctio with ifiitely my derivtives o D d let Dbe rbitrry poit (this ssumptio c be wekeed to sy tht f : D R hs (+) derivtives roud ). The, for ech D (or for ech i tht eighborhood of where f hs (+) derivtives) we hve: ( ) f '( ) f ''( ) f ( ) f ( ) f ( ) R ( ), where the remider (the!!! error) (7) R ( ) is (8) ( ) f () c R ( )!, where c is some poit betwee d. NOTE: This theorem is importt for two resos: It sttes tht the Tylor series eists whe f( ) is lyticl o D (or eve whe it hs t lest (+) roud =); derivtives More importtly, it gives estimte for the error mde whe pproimtig the fuctio oly with fiite umber of terms i the epsio. I prctice, we cot clculte ifiity umber of terms i 8

29 the Tylor epsio, d we wt to kow wht is the size of the error mde. However, ote tht the ew costt c is ot kow, ll we kow is tht it is costt betwee d (remember the me vlue theorem). We will hve to del with this. Sometimes it is coveiet to fid the rdius of covergece of the Tylor series i (7) R, d the we c sy tht R c R which c be useful. The proof of Tylor s theorem is bsed o the Rolle s theorem (try for = followig book) do if time --. Emples: Fid the McLuri series bove d gree with previous formuls. Memorize. Fid McLuri series for p f ( ), f ( ) l, f ( ), p R. Emphsize tht they eed to memorize the formul bove, uderstd d memorize (5), (7) d (8) d memorize them. Homework: Wht left from bove plus,4,7,,9,,3,30, The Tylor Approimtio Theorem: Tylor s theorem is very importt theorem i Clculus. Note tht you met it i vrious prticulr forms before. f ( ) f ( ) For emple, the defiitio of the derivtive f '( ) lim Tylor epsio roud d see if it is true). is prticulr form (replce f( ) with its I this sectio we focus o lyzig the error i Tylor (or McLuri) pproimtio for lyticl fuctio f( ). Wht Tylor s theorem sys is tht y lyticl fuctio is ifiite polyomil t y poit i its domi. Prcticlly, it c be pproimted by (fiite) polyomil t y poits i its domi. Lookig t the formul for the error ( ) f () c R ( )! mde i this pproimtio, we ifer: The error R ( ) decreses by icresig (tkig more terms i the pproimtio) ote tht pproimtig fuctio by lie t poit is rough (this is the pproimtio by lie with the slope f '( ) t poit: f ( ) f ( ) f '( ) ) ; 9

30 The error R ( ) is reltively smll whe is er, but it gets lrge s goes fr from (look t Figure 4 i book, try emple si( ) tkig more d more terms, look how the error is smll er =0 but 3! 5! 7! lrger d lrger wy from 0, i other words it is O.K. to cosider this pproimtio whe clcultig si(0.5) sy, but ot O.K. whe clcultig 5 si! The error becomes bigger s f( ) is less smooth (lrger vlues of the derivtive) over its domi. Emple: Approimte Solutio: 0.7 e with error less th 0.00 (do ot just plug 0.7 e i your clcultor to get the result). Sice 0.7 is close to 0, we c use the McLuri pproimtio: e 3.! 3!! The mi questio is : how lrge should be such tht R ( ) 0.00? Sice ( ) c f () c e R ( ) ( )! ( )! c c e e ( )! ( )! R (0.7) 0.7. However, remember tht c c 0.7 e e 3 R 3 (0.7) 0.00.! By tril d error, we see tht 6, for which we fid T (0.7) which gives pproimtio of e error less th Homework: 5,6,5,9-8 (choose 5), 3,38,39,4,43,56,6 CHAPTER REVIEW: ALL with 30